Force = Rate of change of momentum

Velocity with which a sand particle strikes the bottom of hour-glass,

$v=u+g t$

$\Rightarrow v=0+10\times 2=20 {\mathrm{ms}}^{-1}$

Change in momentum of particle

$=p_f-p_i=0-mv$

$=-0.2\times {10}^{-3}\times 20$

Momentum imparted to base by the particle $=4\times {10}^{-3} \mathrm{kg}·{\mathrm{ms}}^{-1}$

Total change of momentum imparted per second by all 100 particles

$\begin{array}{l}=4\times {10}^{-8} \mathrm{kg}·{\mathrm{ms}}^{-1}\times 100 {\mathrm{s}}^{-1}\\ =0.4 \mathrm{kg}·{\mathrm{ms}}^{-1}\end{array}$

So, force on bottom $=0.4 \mathrm{N}$