Tilted Drum

We note that the volume of water spilled out $\Delta V$ is a half of the volume of a cylinder same radius $r$ as the drum but with a height of $2r \tan 10^\circ$ . In equation,

$\begin{aligned} \Delta V & = \frac {2r\tan 10^\circ}{2h} V & \small \blue{\text{where }h \text{ and }V \text{ are the height and volume of the drum.}} \\ & = \tan 10^\circ \sqrt{\frac {1000V^3}{\pi h^3}} & \small \blue{\text{Note that }r=\sqrt{\frac {1000V}{\pi h}}} \\ & = \tan 10^\circ \sqrt{\frac {1000\cdot 210^3}{\pi \cdot 90^3}} & \small \blue{\text{and }V = 210 \text{ and }h=90} \\ & \approx 11.2 \end{aligned}$

Therefore the remaining volume of water $V - \Delta V = 210 - 11.2 \approx \boxed{199} \text{ L}$ .

$Height~of~water~at~the~other~end~h_1=90-2r*Tan(10)~cm.\\$

$r=\sqrt{\dfrac{210000}{90*\pi}}=\sqrt{\dfrac{7000}{3*\pi}} ~cm.$
$So~water ~left=\pi*r^2*\dfrac{90+h_1} 2=1/2*\pi*7000/(3*\pi)*(180-2*\sqrt{7000/(3*\pi)}*Tan(10))~ cm^3\approx 199~ L\\$

V=pi (R^2)[H-2R Tan(x)+R Tan(x)], H=90 R=sqrt[0.21 10^6/(H*pi)]=27.253, x=10 degree V=198.78, Answer=199