Here's the conversion. I will write Gaussian quantities as
$$\hat{φ}, 𝐚, 𝐛, 𝐞, 𝐝, 𝐡, 𝐣, \hat{ρ}, q, \hat{ε}, \hat{μ}$$
and the corresponding SI quantities as
$$φ, 𝐀, 𝐁, 𝐄, 𝐃, 𝐇, 𝐉, ρ, e, ε, μ.$$
Kinematic and mechanical quantities $𝐯$ (velocity), $𝐩$ (momentum), $m$ (rest mass), $M$ (moving mass), $E$ (energy), $𝐅$ (force), $P$ (power) do not differ between the two. I will also use $𝔏$ to denote the Maxwell-Lorentz Lagrangian density, which is also the same for the two.
Corresponding to the equations in Gaussian form:
$$
𝐛 = ∇×𝐚, \hspace 1 em 𝐞 = -∇\hat{φ} - \frac{1}{c} \frac{∂𝐚}{∂t}, \hspace 1em ∇·𝐛 = 0, \hspace 1em ∇×𝐞 + \frac{1}{c}\frac{∂𝐛}{∂t} = 𝟎, \\
∇·𝐝 = 4π\hat{ρ}, \hspace 1em ∇×𝐡 - \frac{1}{c}\frac{∂𝐝}{∂t} = \frac{4π𝐣}{c}, \hspace 1em ∇·𝐣 + \frac{∂\hat{ρ}}{∂t} = 0, \\
𝔏 = \frac{|𝐞|^2 - |𝐛|^2}{8π}, \hspace 1em 𝐝 = \hat{ε}𝐞, \hspace 1em 𝐛 = \hat{μ}𝐡, \\
𝐅 = q\left(𝐞 + \frac{𝐯×𝐛}{c}\right), \hspace 1em P = q𝐯·𝐞
$$
are the equations in SI form:
$$
𝐁 = ∇×𝐀, \hspace 1 em 𝐄 = -∇φ - \frac{∂𝐀}{∂t}, \hspace 1em ∇·𝐁 = 0, \hspace 1em ∇×𝐄 + \frac{∂𝐁}{∂t} = 𝟎, \\
∇·𝐃 = ρ, \hspace 1em ∇×𝐇 - \frac{∂𝐃}{∂t} = 𝐉, \hspace 1em ∇·𝐉 + \frac{∂ρ}{∂t} = 0, \\
𝔏 = \frac{ε_0|𝐄|^2}{2} - \frac{|𝐁|^2}{2μ_0}, \hspace 1em 𝐃 = ε𝐄, \hspace 1em 𝐁 = μ𝐇, \\
𝐅 = e\left(𝐄 + 𝐯×𝐁\right), \hspace 1em P = e𝐯·𝐄,
$$
with $c = 1/\sqrt{ε_0μ_0}$.
The conversions are:
$$
(\hat{φ}, 𝐚, 𝐛, 𝐞) = \sqrt{4πε_0} (φ, 𝐀c, 𝐁c, 𝐄), \\
(𝐝, 𝐡, 𝐣, \hat{ρ}, q) = \frac{1}{\sqrt{4πε_0}}\left(4π𝐃, \frac{4π𝐇}{c}, 𝐉, ρ, e\right), \\
(\hat{ε}, \hat{μ}) = \left(\frac{ε}{ε_0}, \frac{μ}{μ_0}\right).$$
In the reverse direction, they are
$$
(φ, 𝐀, 𝐁, 𝐄) = \frac{1}{\sqrt{4πε_0}}\left(\hat{φ}, \frac{𝐚}{c}, \frac{𝐛}{c}, 𝐞\right), \\
(𝐃, 𝐇, 𝐉, ρ, e) = \sqrt{4πε_0}\left(\frac{𝐝}{4π}, \frac{c𝐡}{4π}, 𝐣, \hat{ρ}, q\right), \\
(ε, μ) = (ε_0 \hat{ε}, μ_0 \hat{μ}).
$$
There is no direct mention of $ε_0$ or $μ_0$ in the Gaussian version, only of $c$; while in the SI version, there is no direct mention of $c$, only of $ε_0$ and $μ_0$ and it's confined to the (vacuum version of the) constitutive relations and Lagrangian density, nowhere else. All the other equations are $c$-independent and (in fact) non-metrical. They live at a deeper level of geometry where there is no concept of a metric, parallelism, orthogonality, congruence, speed, or any distinction between space-like and time-like.
If the dimensions are denoted $Q$, $P$, $M$, $L$ and $T$, respectively for electric charge/flux, magnetic charge/flux, mass, length and time duration, then the dimensions for the various quantities in SI will be
$$[\left(φ, 𝐀, 𝐁, 𝐄, 𝐃, 𝐇, 𝐉, ρ, e, ε, μ\right)] = \left(\frac{P}{T}, \frac{P}{L}, \frac{P}{L^2}, \frac{P}{LT}, \frac{Q}{L^2}, \frac{Q}{LT}, \frac{Q}{L^2T}, \frac{Q}{L^3}, Q, \frac{QT}{PL}, \frac{PT}{QL}\right).$$
The dimensions of $P$ and $Q$ are conjugate and multiply out to those of action:
$$PQ = \frac{ML^2}{T}.$$
From this, you can determine what the dimensions for the Gaussian quantities are:
$$[\left(\hat{φ}, 𝐚, 𝐛, 𝐞, 𝐝, 𝐡, 𝐣, \hat{ρ}, q, \hat{ε}, \hat{μ}\right)] = \left(\frac{G}{L}, \frac{G}{L}, \frac{G}{L^2}, \frac{G}{L^2}, \frac{G}{L^2}, \frac{G}{L^2}, \frac{G}{L^2T}, \frac{G}{L^3}, G, 1, 1\right), \hspace 1em G = \sqrt{\frac{ML^3}{T^2}}.$$
They are, needless to say, awkward - square roots of dimensions involving $M$, $L$ and $T$. But at least they got rid of the $P$'s and $Q$'s. You might say that they paid no mind to the $P$'s and $Q$'s.
Finally, the kinematic and mechanical dimensions are independent of the system:
$$[(𝐅, P, E, 𝐩, m, M, c, 𝐯, 𝔏)] = \left(\frac{ML}{T^2}, \frac{ML^2}{T^3}, \frac{ML^2}{T^2}, \frac{ML}{T}, M, M, \frac{L}{T}, \frac{L}{T}, \frac{M}{LT^2}\right).
$$
