Answer
$t=0.34s$
Work Step by Step
1) We know the block's initial speed $v_0=0, g=9.8m/s^2$ and it has fallen for time $t$
So the speed of the block just before the collision is $$v=v_0+gt=9.8t$$
The distance reached by the block just before the collision is $$s=\Big(\frac{v+v_0}{2}\Big)t=4.9t^2 (1)$$
2) We assume no external force, so the total linear momentum is conserved. We also assume upward to be the positive direction. Therefore,
$$m_{bul}\vec{v_{bul, 0}}+m_{blo}\vec{v_{blo, 0}}=(m_{bul}+m_{blo})\vec{v_f}$$ $$(0.015\times810)+(1.8\times(-9.8t))=(0.015+1.8)\vec{v_f}$$ $$12.15-17.64t=1.815\vec{v_f}$$ $$\vec{v_f}=6.69-9.72t$$
The block rises upward after the collision, so $v_f=(6.69-9.72t)m/s$
3) It has initial velocity $v_0=(6.69-9.72t)m/s$, $v_f=0$ and $g=-9.8m/s^2$. The same distance reached by the block is $$s=\frac{v_f^2-v_0^2}{2g}=\frac{v_0^2-v_f^2}{19.6}=\frac{44.75-130.05t+94.48t^2}{19.6}$$ $$s=2.28-6.64t+4.82t^2 (2)$$
Combine (1) and (2), $$2.28-6.64t+4.82t^2=4.9t^2$$ $$0.08t^2+6.64t-2.28=0$$ $$t=0.34s$$
