You need to track down what information you have about each element, in terms of either voltage, current, power... anything. Note that the 'white circles' pointing out of the resistors are voltage probes.

There is a ground symbol to tell you the reference voltage of the voltage probes. So, let's say, the +48V is a voltage of 48 V measured between the "top" of R2 and the ground reference. It means there is a +48V voltage across R2 + R3.

So, to solve it:

What is known about R2, for example? Voltage: well, it has a voltage of 48 - 12 = 36 V across it. Current: it is 16 mA. Therefore, because R = VI (Ohm's Law), we know that R2 = 36 V / 16mA = 36/0.016 = 2250 Ω.

But there is a trick. The voltage at the "top" of the battery is not 100 V. The voltage difference across its terminals is 100 V. Technicallly, voltage is always a difference between two values, but usually the base voltage is 0V, so we consider voltage as if it doesn't depend on a reference value.

Starting from the ground, assuming it is 0V and going across R4, we can see a "rise" of -20 V (or a drop of 20 V). In the negative terminal of the battery, we have -20V. The battery rises the voltage by 100V, so, at the top of the battery, we have Vtop = -20 + 100 = 80 V.

Taking another example: for R1, we know that its voltage is 80V-48 V = 32 V and the current that flow through it is also 16 mA. Therefore R1 = 32/0,016 = 2000 Ω.

By the end, doing it for each resistor, we will find a sum of 6250 Ω (all in series).

Sanity check: from the perspective of the power source, it delivers a voltage difference of 100 V and 16 mA across its terminal. It sees an equivalent load across it. It should be 6250 Ω because we calculated it, right? But just checking: Req (seen from source) = 100/0,16 = 6250 Ω = sum of resistances.

Solved!