Expected values (almost) always correspond to sums or integrals, depending on whether the random variable is discrete or continuous. In this case, the random variable (the radius) is continuous--it can take on any value between 2 and 4--so it's going to be an integral. The form of the integral is
$$\mathbb E[g(X)] = \int_{-\infty}^{\infty} g(x) f(x) \; dx,$$
where $g(X)$ is an arbitrary function of the random variable $X$ and $f(x)$ is the density function. In this case, $g(x)$ is the formula for the volume of a sphere,
$$g(x) = \frac43 \pi x^3,$$
and $f(x) = \frac12$ between 2 and 4, as you noted, and zero everywhere else. So the expected value is going to be
$$\mathbb E[g(X)] = \int_2^4 \frac43 \pi x^3 \times \frac12 \; dx,$$
which you should be able to solve.
There is nothing wrong with your answer and it does not constradicts the question.
Note when $\arcsin (V)$ in part 2, is not the same $\Theta$ YOU used in the integral.
Note $Y=y, X=x$ and $Y=-y, X=x$ gives the same value of $\arcsin(V)$, i.e.
$\arcsin(x/\sqrt{x^2+y^2}) = \arcsin(x/\sqrt{x^2+(-y)^2})$ but $X=x$ and $Y=y$ and $Y=y, X=-x$ gives different value of $\Theta$ in your parametrisation.
e.g.
$y = 1, x = 1$ gives $\Theta = \pi/4$ but $y=-1, x=1$ gives $\Theta = \pi/4 $
Explicitly
$\arcsin (V) = \Theta$ for $\Theta \in [0,\pi/2]$
$\arcsin (V) = -\Theta + \pi$ for $\Theta \in [3\pi/2, 2\pi]$
$\arcsin (V) = \Theta - \pi$ for $\Theta \in (\pi/2, \pi)$
$\arcsin (V) = -\Theta +2\pi $ for $\Theta \in [3\pi/2, 2\pi)$
so $\arcsin (V)$ is a uniform distribution on $(-\pi/2, \pi/2)$
Best Answer
Observe that if $R$ is lognormally distributed with parameters $\mu$ and $\sigma$, then $R = e^X$, where $X$ is normally distributed with mean $\mu$ and standard deviation $\sigma$.
Therefore, for any nonzero real scalar $k$, the variable $R^k = e^{kX}$ is also lognormal because $kX$ is normal with mean $k \mu$ and standard deviation $k \sigma$.
In other words, because scale transformations of a normally distributed random variable are also normally distributed, and because lognormal random variables are exponential transformations of normal random variables, then nonzero powers of lognormal random variables are lognormal.
Similarly, location transformations of a normally distributed random variable are also normally distributed, so multiplication of a lognormal random variable by a positive constant also results in a lognormal random variable. That is to say,
$$c R^k = c e^{kX} = e^{k X + \log c},$$ so the new parameters are $k \mu + \log c$ and $k \sigma$. In your case, $k = 3$ and $c = \frac{4}{3}\pi$.