The maximum velocity of the photoelectrons emitted from the surface...

According to Einstein's photoelectric effect,

K{E}_\max = \dfrac{hc}{\lambda} - ϕ

The incident energy of the photon is ${E}_{1} = \dfrac{1}{2}m{v}^{2} = hn - ϕ . . . \left(1\right)$

If incident frequency increased to 3n

${E}_{2} = \dfrac{1}{2}m{v}_{2}^{2} = 3hn - ϕ$

$= 3\left(hn - ϕ\right) + 2ϕ$

${E}_{2} = 3{E}_{1} + 2ϕ$

$⇒ \dfrac{1}{2}m{v}_{2}^{2} = \dfrac{3}{2}m{v}^{2} + 2ϕ$

$⇒ {v}_{2}^{2} = 3{v}^{2} + \dfrac{4ϕ}{m}$

$⇒ {v}_{2} > v\sqrt{3}$