$$
r=frac{1}{B}=sqrt{frac{2mV}{q}}
$$

so if all other parameters are kept constant the only parameter that is unknown is the particle’s mass and if we know the radius of the path by detecting the place where the particle hits the screen, we can calculate the mass.

$$
r=frac{1}{B}sqrt{frac{2mV}{q}}
$$

we see that

$$
rpropto frac{sqrt{m}}{B}
$$

So if we want to keep it in a given range of values, so more or less constant it has to hold that

$$
Bproptosqrt{m}
$$

So the ratio between the fields is then given as

$$
boxed{frac{B_2}{B_1}=frac{sqrt{m_2}}{sqrt{m_1}}}
$$

If the mass increases 100 times the magnetic field has ti increase 10 times.

$$
qvB=frac{mv^2}{r}
$$

From here, we can express the radius to have that

$$
r=frac{mv}{qB}=frac{1.67times 10^{-27}times 4.2times 10^4}{1.6times 10^{-19}times 1.2}
$$

Finally, we have that

$$
boxed{r=3.65times 10^{-4}textrm{ m}}
$$

$$
qvB=frac{mv^2}{r}
$$

However, we have to find the speed first. The final speed of the oxygen atoms is the consequence of the acceleration in the electric field which is expressed via the conservation of energy

$$
qV=frac{mv^2}{2}
$$

where we can express the square of the speed as

$$
v^2=frac{2qV}{m}
$$

Now, we will square our first equation and plug in the last expression into it

$$
q^2v^2B^2=frac{m^2v^4}{r^2}
$$

$$
q^2B^2=frac{m^2v^2}{r^2}=frac{m^2}{r^2}frac{2qV}{m}
$$

After some simple algebra one gets that

$$
qB^2=frac{2mV}{r^2}
$$

So we can express the mass as

$$
m=frac{qr^2B^2}{2V}=frac{2times1.6times 10^{-19}times 0.075^2times 0.083^2}{2times 232}
$$

Finally, we have that

$$
boxed{m=2.7times 10^{-26}textrm{ kg}}
$$