This does work out to be correct, but there's a pitfall hiding in the question in the idea of what the "average speed" is, and (IMO) the subtleties involved make it a bad fit for a ninth-grade math class.
To be a bit more realistic, let's assume that the nozzle is a frustum with height $h$, a radius $r_0$ at the base and radius $\epsilon r_0$ (with $\epsilon \ll 1$) at the tip. Assuming the frustum tapers uniformly, the radius of the cone at a distance $x$ from the base would be $r = r_0 + xr_0(\epsilon - 1)/h $, and the speed of the water a distance $x$ from the base would be $$ v = v_0 \frac{r_0^2}{(r_0 + x(\epsilon r_0 - r_0)/h)^2} = \frac{v_0}{(1 + x (\epsilon-1)/h)^2}. $$
This means that we can calculate the distance-averaged speed of the water over the length of the nozzle. This would be what we would get if we stuck a large number of evenly-spaced flow metersflowmeters along the length of the nozzle and then averaged their readings. It works out to be $$ \langle v \rangle_x = \frac{v_0}{h} \int_0^h \frac{dx}{(1 + x (\epsilon-1)/h)^2} = \frac{v_0}{h} \left( \frac{h}{\epsilon} \right) = \frac{v_0}{\epsilon}. $$ We can see that this is not equal to $3 v_0$ as your argument would predict.
OnBut wait! Maybe if we take the other handaverage over the volume of the nozzle, things will work better. We can accomplish this by "weighting" the above integral by the cross-sectional area of the nozzle. This does work out; we have $$ \langle v \rangle_V = \frac{v_0}{V} \int_0^h \frac{\pi r_0^2 (1 + x (\epsilon-1)/h)^2 dx}{(1 + x (\epsilon-1)/h)^2} = \frac{\pi h v_0 r_0^2}{\frac{1}{3} \pi h r_0^2(1 + \epsilon + \epsilon^2)} = \frac{3 v_0}{1 + \epsilon + \epsilon^2}. $$ which does in fact approach $3 v_0$ as $\epsilon \to 0$.
Alternately, we can also try to calculate the average speed of a water molecule as it passes through the nozzle. The time taken for the water molecule to pass through this nozzle will be $$ \Delta t = \int_{x=0}^{x=h} dt = \int_0^h \frac{dx}{v} = \frac{h}{3 v_0} (1 + \epsilon + \epsilon^2) $$ and the time-averaged speed of the water molecule will then be the distance traveled $h$ divided by $\delta t$, or $$ \langle v \rangle_t = \frac{3 v_0}{1 + \epsilon + \epsilon^2} $$$$ \langle v \rangle_t = \frac{3 v_0}{1 + \epsilon + \epsilon^2}, $$ which does in fact approach $3 v_0$ as $\epsilon \to 0$as we found above by the volume average.
So if you define "average speed" to be distance traveled over time taken, then your answer is correct. To be fair, this is how "average speed" is defined in physics, so it's technically correct. But it's also true that the speed of the water gets very very very large for $\epsilon \ll 1$; for a proper cone ($\epsilon = 0$), it is moving infinitely fast as it exits the nozzle. You might get a student asking you "if the water is moving infinitely fast as it leaves the cone, how can the average speed be finite?" I'm not sure that there's a good answer to this question without breaking out the calculus like I did above, and presumably ninth-graders aren't ready for that.
(This idea that an infinite speed at one point in a trip is compatible with a finite average speed over that trip is more clearly (?) illustrated in the old chestnut of a brain-teaser. A person needs to drive a 10-mile journey. They run into traffic over the first 5 miles and only drive at 30 miles per hour over that distance. How fast must they drive for the remaining 5 miles to average 60 miles per hour over the whole trip? The intuitive answer of 90 mph is, of course, wrong, but very smart people have reportedly fallen into that trap.)