First let's use a conceptual approach, as this multiple choice problem is answerable in that way.
$K_\mathrm{eq}$ cannot be negative. Therefore $\mathrm{A}$ is incorrect.
$K_{\mathrm{eq}}$ is greater than one when the $\ce{left -> right}$ reaction goes farther; it is less than one when the reverse reaction goes farther (by going farther I mean end in greater concentration at equilibrium).
The
$$\ce{H2 + I2 <=> 2HI}$$
reaction has $K_{\mathrm{eq}}$ over $1$. Therefore the
$$\ce{H2 + I2 -> 2HI}$$
happens faster than the
\begin{align}
&&\ce{H2 + I2 &<- 2HI}\\
\text{or}&&
\ce{2HI &-> H2 + I2}.
\end{align}
So in
$$\ce{2HI <=> H2 + I2}$$
or the reverse, the $\ce{left->right}$ reaction occurs slower; the $K_{\mathrm{eq}}$ must be less than $1$. $\mathrm{B}$ is the only answer.
Another way to approach the problem is using algebraic reasoning.
Initially,
$$K_{eq} = \frac{\ce{[HI]^2}}{\ce{[H2][I2]}} = 50.$$
Then, when you reverse the reaction,
$$K_{eq} = \frac{\ce{[H2][I2]}}{\ce{[HI]^2}}
\frac{1}{\left(\frac{\ce{[HI]^2}}{\ce{[H2][I2]}}\right)}= \frac{1}{50.0} = 0.0200.$$
In general when you reverse the equation
$$K_{\text{eq (new)}} = \frac{\text{products (new)}}{\text{reactants (new)}} = \frac{\text{reactants (old)}}{\text{products (old)}} = \frac{1}{K_{\text{eq (old)}}}.$$