Answer
$\textbf{(a)}$ $t=2.95\times10^{-9}\text{ s}.$
$\textbf{(b)}$ $d_v=2.31\text{ cm}.$
Work Step by Step
$\textbf{(a)}$ In the horizontal direction, the particle moves uniformly so the required time is given by
$$t=\frac{d_h}{v_h}=\frac{6.20\text{ cm}}{2.10\times10^9\text{ cm/s}}=\frac{6.20}{2.10}\times 10^{-9}\text{ s}=2.95\times10^{-9}\text{ s}.$$
$\textbf{(b)}$ In the vertical direction the electron moves with uniform acceleration without the initial velocity so the displacement in the vertical direction is given by:
$$d_v=\frac{1}{2}a_v t^2=\frac{1}{2}\cdot5.30\times10^{17}\text{ cm/s}^2\cdot(2.95\times 10^{-9}\text{ s})^2=\\
\frac{1}{2}\cdot5.30\cdot2.95^2\times10^{17-18}\text{ cm}=23.10\times10^{-1}\text{ cm}=2.31\text{ cm}.$$
