d
(D)
Let final pressure inside tube be $P$
Apply $P_1 V_1=P_2 V_2$
$10^5 \times 20= P \times(20- h ) \ldots(1)$
Pressure Equation
$P = P _0+\rho g \frac{(10- h )}{100}$
$P =10^5+10^3 \times 10 \times \frac{(10- h )}{100}$
$P =10^5\left(1+\frac{10- h }{10^3}\right) \ldots(2)$
Substitute $(2)$ in $(1)$
$10^5 \times 20=10^5\left(1+\frac{10-h}{10^3}\right)(20- h )$
Solve for $h$
$h =0.2 \,cm$