When light of wavelength 248nm falls on a metal of threshold energy...

Energy incident $= \dfrac{hc}{\lambda}$

$= \dfrac{6 . 63 \times {10}^{- 34} \times 3 . 0 \times {10}^{8}}{248 \times {10}^{- 9} \times 1 . 6 \times {10}^{- 19}}eV$

$= \dfrac{6 . 63 \times 3 \times 100}{248 \times 1 . 6}$

$= 0 . 05eV \times 100 = 5eV$

Now using

$E = ϕ + K . E$

$5 = 3 + K . E .$

$K . E . = 2eV = 3 . 2 \times {10}^{- 19}J$

for de Broglie wavelength $\lambda = \dfrac{h}{mv}$

$K . E . = \dfrac{1}{2}{mv}^{2}$

so $v = \sqrt{\dfrac{2K . E .}{m}}$

hence $\lambda = \dfrac{h}{\sqrt{2K . E . \times m}}$

$= \dfrac{6 . 63 \times {10}^{- 34}}{\sqrt{2 \times 3 . 2 \times {10}^{- 19} \times 9 . 1 \times {10}^{- 31}}}$

$= \dfrac{6 . 63}{7 . 6} \times \dfrac{{10}^{- 34}}{{10}^{- 25}} = \dfrac{66 . 3 \times {10}^{- 10}m}{7 . 6}$

$= 8 . 72 \times {10}^{- 10}m$

$\approx 9 \times {10}^{- 10}m$

$= 9\overset{o}{A}$