Use Einstein’s photoelectric equation

$h\nu = h\nu_0 + \frac{1}{2}mv^2$

Hence,

$\frac{1}{2}mv^2 = h(\nu-\nu_0)$

Calculate frequency of incident radiation

Given wavelength,

$\lambda = 500\,\text{nm} = 500\times 10^{-9}\,\text{m}$

Frequency is

$\nu = \frac{c}{\lambda} = \frac{3\times 10^8}{500\times 10^{-9}}$

$\nu = 6\times 10^{14}\,\text{Hz}$

Given threshold frequency

$\nu_0 = 4.3\times 10^{14}\,\text{Hz}$

Therefore,

$\nu-\nu_0 = (6.0-4.3)\times 10^{14} = 1.7\times 10^{14}\,\text{Hz}$

Compute kinetic energy of emitted electron

$K.E. = h(\nu-\nu_0)$

$= 6.63\times 10^{-34}\times 1.7\times 10^{14}$

$= 11.271\times 10^{-20}\,\text{J}$

$= 1.1271\times 10^{-19}\,\text{J}$

Use

$\frac{1}{2}mv^2 = 1.1271\times 10^{-19}$

with $m = 9.0\times 10^{-31}\,\text{kg}$:

$v^2 = \frac{2\times 1.1271\times 10^{-19}}{9.0\times 10^{-31}}$

$v^2 = \frac{2.2542\times 10^{-19}}{9.0\times 10^{-31}}$

$v^2 = 2.5047\times 10^{11}$

Find velocity

$v = \sqrt{2.5047\times 10^{11}}$

$v \approx 5.0\times 10^5\,\text{m s}^{-1}$

Nearest integer

The blank is in the form $\underline{\hspace{2cm}}\times 10^5\,\text{m s}^{-1}$, so the required integer is

$\boxed{5}$