Standard reduction electrode potentials of three metals A,B and C are $+0.5V,-3.0V\&-1.2V$ respectively.
The decreasing order of reducing powers of these metals is:

The correct option is D $B>C>A$
Higher the standard reduction potential (SRP) of the metal, more is the tendency of that metal to get reduced, and hence lower is its reducing power.
Order of SRP:
$A>C>B$
So, the order of reducing power is $B>C>A$.
Hence, Option (d) is correct.

Theory: $\text{Electrochemical Series}$

$\text{Half-Reaction}$	Standard Reduction Potential $E^{\circ }$ (at 298 K)
$F_2\left(g\right)+2e^{-}\to 2F^{-}\left(aq\right)$	$+2.87V$
$Mn{O}_4\left(aq\right)+8H^{+}\left(aq\right)+5e^{-}\to M{n}^{2+}\left(aq\right)+4H_{2}O\left(l\right)$	$+1.51V$
$A{u}^{3+}\left(aq\right)+3e^{-}\to Au\left(s\right)$	$+1.4V$
$2H{g}^{2+}\left(aq\right)+2e^{-}\to H{g}^{2+}\left(aq\right)$	$+0.92V$
$A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$	$+0.8V$
$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)$	$+0.34V$
$2H^{+}\left(aq\right)+2e^{-}\to H_2\left(g\right)$	$0V$
$P{b}^{2+}\left(aq\right)+2e^{-}\to Pb\left(s\right)$	$-0.13V$
$S{n}^{2+}\left(aq\right)+2e^{-}\to Sn\left(s\right)$	$-0.14V$
$Ni2+\left(aq\right)+2e^{-}\to Ni\left(s\right)$	$-0.25V$
$F{e}^{2+}\left(aq\right)+2e^{-}\to Fe\left(s\right)$	$-0.44V$
$Z{n}^{2+}\left(aq\right)+2e^{-}\to Zn\left(s\right)$	$-0.76V$
$A{l}^{3+}\left(aq\right)+3e^{-}\to Al\left(s\right)$	$-1.66V$
$M{g}^{2+}\left(aq\right)+2e^{-}\to Mg\left(s\right)$	$-2.36V$
$N{a}^{+}\left(aq\right)+e^{-}\to Na\left(s\right)$	$-2.71V$
$C{a}^{2+}\left(aq\right)+2e^{-}\to Ca\left(s\right)$	$-2.87V$
$K^{+}\left(aq\right)+e^{-}\to K\left(s\right)$	$-2.93V$
$L{i}^{+}\left(aq\right)+e^{-}\to Li\left(s\right)$	$-3.05V$

Down the table, as the SRP decreases, the reducing power increases.