How to solve this physics problem about electric current?

Use ohm's law and Kirchhoff's law.

If I through the 300 ohm R (i.e. R300) is 1A then the current through R100 is 3x as large since R100 is 3x smaller. (ohms law). So I through R100 is 3A.

The current through R25 then is 3A + 1A = 4A (Kirchhoff)

Ok, R300 and R100 are in parallel, their total R = (R300 R100) / (R300 + R100) = (100 300)/400 = 75 ohm.

We now have R25 + R75 in parallel with R200. So R25 + R75 = 100 ohm.
The current through this 100 ohm resistor we found to be 4A. (R25 and R75 are in series so the current remains 4A since charge must be conserved)
R200 is twice the value of the 100 ohm resistor so the current is half. The current through R200 = 2A.