Answer
${\bf 8.6 }\;\rm mm $
Work Step by Step
We have here a circular aperture experiment.
We know that the width of the central maxima in such an experiment is given by
$$w=\dfrac{2.44\lambda L}{D}\tag 1$$
where $L$ is the distance between the circular aperture and the screen, $D$ is the diameter of the aperture, and $\lambda$ is the wavelength of the electrons which we can find by using de Broglie wavelength's formula.
$$\lambda=\dfrac{h}{p}=\dfrac{h}{mv} $$
Plug into (1);
$$w=\dfrac{2.44 Lh }{mvD}\tag 2$$
Now the only unknown in (2) is the speed of the electrons $v$.
We can find it by recalling the kinematic energy of
$$K_f=\frac{1}{2}mv^2$$
So the speed of the electron is given by
$$v=\sqrt{\dfrac{2K_f}{m}}= \sqrt{\dfrac{2 eV}{m}}$$
Plug the known;
$$v=\sqrt{\dfrac{2(1.6\times 10^{-19})(250)}{(9.11\times 10^{-31})}}=\bf 9.371\times 10^6\;\rm m/s$$
Now we can plug all the known into (2) to find the diameter (the width) of the bright spot (central maxima).
$$w=\dfrac{2.44 (1.5)(6.63\times10^{-34})}{(9.11\times 10^{-31})(9.371\times 10^6)(33\times 10^{-9})}=\bf \color{red}{\bf 8.6\times 10^{-3}}\;\rm m $$
