Given: $v=3\times10^7$ m/s, plate separation $d=2$ cm $=0.02$ m, plate length $L=3$ cm $=0.03$ m.

Beam enters symmetrically and just strikes the end of plate $P_2'$, so vertical deflection $y=d/2=0.01$ m.

Time spent between plates:

$$t=\frac{L}{v}=\frac{0.03}{3\times10^7}=1\times10^{-9}\text{ s}$$

Vertical deflection: $y=\frac12at^2 \Rightarrow a=\dfrac{2y}{t^2}=\dfrac{2\times0.01}{(1\times10^{-9})^2}=2\times10^{16}\text{ m/s}^2$

Acceleration due to field: $a=\dfrac{eE}{m}=\dfrac{eV}{md}$, so:

$$V=\frac{amd}{e}=\frac{(2\times10^{16})(9.1\times10^{-31})(0.02)}{1.6\times10^{-19}}\approx2275\text{ V}\approx2.28\text{ kV}$$