For neatness and convenience, I'll use the notation 1.6E-19 (for example) instead of 1.6x10⁻¹⁹.

________________________________

Proton's initial kinetic energy K1 = ½mv₀²

= ½ x 1.67E-27 x (1.6E6)²

= 2.138E-15J

The potential energy at the start point P1 = kqQ/r

= 9E9 x 1.6E-19 x -10E-9/0.003

= -4.8E-15 J

The potential energy at the end point P2 = kqQ/r

= 9E9 x 1.6E-19 x -10E-9/0.004

= -3.6E-15 J

Since energy is conserved:

K1 P1 = K2 P2

K2 = K1 P1 - P2

. . .= 2.138E-15 (-4.8E-15) - (-3.6E-15)

. . .= 9.38E-16 J

½mv² = 9.38E-16

v = √[2 x 9.38E-16 / 1.67E-27]

. .= 1.06E6 m/s

__________________________________

Of course the solution can be written much quicker than that. I would usually just write:

½mv₀² kqQ/r1 = ½mv² kqQ/r2

½mv² = ½mv₀² kqQ(1/r1 - 1/r2)

and then substitute values.

Lorem ipsum dolor sit amet, consectetur adipiscing elit,
sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam,
quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor
in reprehenderit in voluptate velit esse cillum dolore eu fugiat.

SHOW ANSWER