An infinitely long conductor $PQR$ is bent to form a right angle as shown. A current $I$ flows through $PQR$. The magnetic field due to this current at the point $M$ is $H_{1}$. Now another infinitely long straight conductor $QS$ is connected at $Q$ so that the current is $I/2$ in $QR$ as well as in $QS$. The current in $PQ$ remaining unchanged. The magnetic field at $M$ is now $H_{2}$. The ratio $H_{1}/H_{2}$ is given by 

At $M$, the horizontal part $PQ$ gives no field because for every current element on $PQ$, the position vector toward $M$ is along the same line, so $d\vec{l} \times \vec{r} = 0$. So initially only the vertical semi-infinite part $QR$ contributes. For a semi-infinite straight wire at perpendicular distance $a$, $$H = \frac{I}{4\pi a}$$ Hence, $$H_1 = \frac{I}{4\pi a}$$ After connecting $QS$, the current in $QR$ becomes $\frac{I}{2}$ and the current in $QS$ is also $\frac{I}{2}$. At $M$, the branch $QS$ also gives no field, since $M$ lies on the same straight line produced through $QS$, so again $d\vec{l} \times \vec{r} = 0$ for that branch. Thus only $QR$ contributes in the new case: $$H_2 = \frac{I/2}{4\pi a} = \frac{I}{8\pi a}$$ Therefore, $$\frac{H_1}{H_2} = \frac{I/(4\pi a)}{I/(8\pi a)} = 2$$ This matches option $D$.