Finding out modulation index and DC offset
The sinusoid test signal is given by
$$
s(t) = V_0\cos(2\pi f_0)
$$
and its Fourier transform by
$$
S(f) = \frac{V_0}{2}\big[\delta(f-f_0) + \delta(f+f_0)\big].
$$
Inserted in $S_\mathrm{AM-DSB-C}$ it yields
$$
S_\mathrm{AM-DSB-C}(f) = \frac{AV_0}{4}\big[\delta(f-f_c-f_0) + \delta(f-f_c+f_0) + \delta(f+f_c-f_0) + \delta(f+f_c+f_0) \big] + \frac {Ac}{2} \big[\delta (f-f_c) + \delta (f+f_c)\big].
$$
Comparing that with the figure given we see that $AV_0/4 = 3$ and $Ac/2 = 4$. Additionaly, $f_0 = 10\text{ kHz, } f_c = 100\text{ kHz}$. Using $m = V_0/c$ and $V_0 = 6\text{ V}$ we find
$$
m = 1.5, c = \frac{V_0}{m}=4\text{ V}.
$$