To find the standard potential of M$$^{3+}$$/M electrode, the following cell is constituted:
Pt | M/M$$^{3+}$$ (0.001 mol L$$^{-1}$$) / Ag$$^+$$ (0.01 mol L$$^{-1}$$) / Ag
The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half-reaction M$$^{3+}$$ + 3e$$^-$$ $$\rightarrow$$ M at 298 K will be:
(Given: $$E^\ominus_{Ag^+/Ag}$$ at 298 K = 0.80 volt)

First, write the two relevant standard reduction half-reactions along with the known standard potential of silver:

$$Ag^+ + e^- \rightarrow Ag, \;\; E^\circ_{Ag^+/Ag}=0.80\;{\rm V}$$

$$M^{3+}+3e^- \rightarrow M, \;\; E^\circ_{M^{3+}/M}=E^\circ_M \;({\rm unknown})$$

To combine these two half-reactions the electrons must balance. Multiplying the silver reaction by 3 gives

$$3Ag^+ + 3e^- \rightarrow 3Ag$$

and adding this to the reduction of $$M^{3+}$$ (written in the reverse direction because it will actually supply electrons) yields the overall spontaneous reaction taking place in the cell:

$$3Ag^+ + M \rightarrow 3Ag + M^{3+}$$

In this overall reaction a total of $$n = 3$$ electrons are transferred.

The cell emf at 298 K is connected to the reaction quotient by the Nernst equation:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q$$

Here,

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$

Because the silver electrode possesses the larger standard potential, it functions as the cathode, while the $$M/M^{3+}$$ electrode acts as the anode. Hence,

$$E^\circ_{\text{cell}} = E^\circ_{Ag^+/Ag} - E^\circ_{M^{3+}/M} = 0.80 - E^\circ_M$$

The reaction quotient $$Q$$ for the overall reaction is constructed from concentrations of ionic species (activities of solids are unity):

$$Q = \frac{[M^{3+}]}{[Ag^+]^3}$$

Given $$[M^{3+}] = 0.001\;{\rm mol\,L^{-1}}$$ and $$[Ag^+] = 0.01\;{\rm mol\,L^{-1}}$$, we have

$$Q = \frac{0.001}{(0.01)^3} = \frac{0.001}{1\times10^{-6}} = 1000$$

$$\log Q = \log(1000) = 3$$

The measured emf of the cell at 298 K is $$E_{\text{cell}} = 0.421\;{\rm V}$$. Substituting all known values into the Nernst equation gives

$$0.421 = (0.80 - E^\circ_M) - \frac{0.0591}{3}\,\times 3$$

Since $$\dfrac{0.0591}{3}\times3 = 0.0591$$, this simplifies to

$$0.421 = 0.80 - E^\circ_M - 0.0591$$

Now bring like terms together:

$$0.421 = 0.7409 - E^\circ_M$$

Rearranging gives the desired standard potential:

$$E^\circ_M = 0.7409 - 0.421 = 0.3199\;{\rm V} \approx 0.32\;{\rm V}$$

Hence, the correct answer is Option C.