Answer is A
$P(X\leq -1)=P(Y\geq2)$
We can compare their values using standard normal distributions:
$$\begin{array}{c|c}
X& Y\\
\mu_X = +1 & \mu_Y = -1\\
{\large\sigma}_{X}{^2}=4&{\large\sigma}_{Y}{^2}=?\\
Z_{X}=\dfrac{X-1}{\sqrt{4}}&Z_{Y}=\dfrac{Y-(-1)}{{\large\sigma}_{Y}}\\
2Z_{X}+1=X&Y={\large\sigma}_{Y}Z_{Y}-1\\
\end{array}$$
$\implies P(2Z_{X}+1\leq -1)=P({\large\sigma}_{Y}Z_{Y}-1\geq 2)$
$\implies P(Z_{X}\leq -1)=P(Z_{Y}\geq \dfrac{3}{{\large\sigma}_{Y}})$
$\implies -(-1)=\dfrac{3}{{\large\sigma}_{Y}}$
$\implies {\large\sigma}_{Y}=3$