How to calculate the energy freed in the reaction: $^{10}_5Be +\space ^2_1H \rightarrow \space^{11}_5B + \space ^1_1H$? I have the following reaction: $^{10}_5\mathrm{Be} +\space ^2_1\mathrm{H} \rightarrow \space^{11}_5\mathrm{B} + \space ^1_1\mathrm{H}$  
And I know that I have to use the formula: $E = \Delta m\cdot c^2 = \Delta m \cdot \frac{931,5MeV}{u}$.
So I just need $\Delta m$ which is equal to:
$\Delta m = m_b - m_a$ where $m_b$ represents the mass "before the reaction" and $m_a$ the mass "after the reaction" so we have:
$m_b = m(^{10}_5\mathrm{Be}) + m(^2_1\mathrm{H})$
$m_a = m(^{11}_5\mathrm{B}) + m(^1_1\mathrm{H})$  
The book which contains this problem contains the following table:
http://i.imgur.com/esoGDVf.png but from this table, I only know $ m(^1_1\mathrm{H})$ and  $m(^2_1\mathrm{H})$ i.e.
$m_b = m(^{10}_5\mathrm{Be}) + 2.01410u$
$m_a = m(^{11}_5\mathrm{B}) + 1.00783u$  
How do I calculate $m(^{10}_5\mathrm{Be})$ and $m(^{11}_5\mathrm{B})$ ?
P.S. I don't know if the tag is correct. The chapter in the book where I found this exercise is called "Basics of nuclear physics".
 A: 
How do I calculate $m\left(^{10}_4\mathrm{Be}\right)$ and $m\left(^{11}_5 \mathrm{B}\right)$?

The masses and various other properties of isotopes are available freely at Wolfram Alpha. They are,


*

*$m\left(^{10}_4\mathrm{Be}\right)=10.013533818u$

*$m\left(^{11}_5 \mathrm{B}\right)=11.009305406u$


where $u$ denotes unified atomic mass units. Notice you are already given the mass number in the superscript of the isotope. As John Rennie noted, the reaction should probably be with $^{9}_4\mathrm{Be}$, 
$$^{9}_4\mathrm{Be} + ^{2}_1 \mathrm{H} \to ^{10}_{5}\mathrm{B} + n$$
in which case the mass is $9.012182201u$.
A: In case you are curious, the other "most likely" interpretation of the incorrect reaction in your problem is
$$
^{10}_5\mathrm{B} +\space ^2_1\mathrm{H} \rightarrow \space^{11}_5\mathrm{B} + \space ^1_1\mathrm{H}
$$
But you still have the quirk that boron-11 (which is stable) is not in your mass table; the nuclear wallet cards give it a mass of $10\text{ u} + 12.051\text{ MeV}/c^2$, which gives this reaction about twice the $Q$-value of neutron production on beryllium suggested by John Rennie and JamalS.  I'd consider the nuclear wallet cards a reliable source (the bound paper version is free! order one) and Wolfram Alpha an amusing convenience.
Beryllium is a very effective neutron reflector, and is often used to increase the neutron flux at reactors.  Boron, on the other hand, is a neutron absorber, and is most commonly used as shielding.  I would expect that if your practical problem is, say, criticality in a reactor, neutron production on beryllium is a more important reaction to consider than proton production on boron.
As to your question itself,

How do I calculate masses like $m(^{11}_5\text{B})$ ?

you determine nuclear masses by measuring the Q-values of reactions like these.  There is a small community of theorists with the tools to predict the masses of light nuclei; those techniques won't be found in any introductory textbook.

In response to a comment: the nuclear wallet cards tabulate the "mass excess," which is the difference (in energy units) between the actual mass of an isotope with mass $A$ and the "naive" mass of $A$ atomic mass units.  The definition of the atomic mass unit is based on carbon-12, so carbon-12 has a mass excess of exactly zero.  I wrote the mass for boron-11 as $10u + 12.051\text{ MeV}/c^2$ because the question was about total mass.  But for determining the reaction energy, you can see immediately that there are 12 nucleons in both the initial and the final state, and so both sides will have an approximate mass of $12\,\text{u} \approx 12\,000\,\text{MeV}$. The only thing to be gained by using the "exact" amu-to-MeV conversion factor $1\,\text{u} = 931.414\,013\,\text{MeV}$ is some opportunities for arithmetic errors.
With this approach the Q-value is 
$$
\begin{array}{rl}
+12.051\text{ MeV} & \text{from boron-10} \\
+13.136\text{ MeV} & \text{from deuterium} \\
-8.668 \text{ MeV} & \text{to boron-11} \\
-7.289 \text{ MeV} & \text{to hydrogen} \\
\hline\\
9.410 \text{ MeV} & \text{to kinetic energy in final state}
\end{array}
$$
I'll leave neutron production on beryllium to you — I get about 4 MeV.
