There are two ways of solving this problem, both having same calculations.

One is by Using Kirchoff's Law and one without. I am telling this without using the Law, because if you understand this, you can do it using the Law.

In the image attached, I have installed the internal resistance outside the ideal battery and limited the battery between the limits E and F. So, I can say my whole battery is between E and F. Now Since the battery is to be charged, it need to be placed in opposite way as the Charger's terminal's are connected with the series [Chargeable batteries need to **reverse the chemical reactions** to be charged]. So the effective Potential Difference will be 110-12 = 98 V. Now, there are two resistance and both in series [your question says so]. The total resistance is 0.5 +x.

Now using the formula V=IR, I given as 5 Amperes [Current to be restricted to 5 A], value of x comes to be 19.1 Ohms.

Please tell me the answer so that I can recheck mistakes.