MTBF calculation with 2 units connected in series
If the MTBF of a device is $N$, then the probability that the device will fail in 1 time unit is $1/N$. With two devices in series with failure probabilities $p_1$ and $p_2$, the system will fail in one time unit if device 1 fails or if device 1 doesn't fail and device 2 fails. The probability of that happening is
$$
p_1 + (1-p_1)p_2=p_1+p_2-p_1p_2
$$
In your problem, $p_1=1/200, p_2=1/300$ so the probability that the system will fail in one time unit is 
$$
(1/200)+(1/300)-(1/200)(1/300)=499/60000\approx0.00832
$$
and so the MTBF of the whole system is $1/0.00832\approx120.24$. Your friend is right: none of the proposed solutions are correct.