Galvanic cells

The couple with the higher reduction potential acts as the cathode and the one with the lower reduction potential as the anode.
Cathode (reduction): $Sn^{4+} + 2e^- \rightarrow Sn^{2+}$, $E^\circ = +0.15$ V
Anode (oxidation): $Cr \rightarrow Cr^{3+} + 3e^-$, $E^\circ_{Cr^{3+}/Cr} = -0.74$ V
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
$E^\circ_{cell} = 0.15 - (-0.74)$
$E^\circ_{cell} = +0.89$ V

The basic relation is: $E_{cell} = E_{cathode}(\text{reduction}) - E_{anode}(\text{reduction})$
Oxidation potential of an electrode = −(its reduction potential).
Replacing the anode term: $-E_{anode}(\text{red}) = +E_{anode}(\text{oxid})$, so $E_{cell} = E_{anode}(\text{oxid}) + E_{cathode}(\text{red})$. This is relation (b).
Replacing the cathode term as well: $E_{cathode}(\text{red}) = -E_{cathode}(\text{oxid})$, so $E_{cell} = E_{anode}(\text{oxid}) - E_{cathode}(\text{oxid})$. This is relation (d).
Relations (a) and (c) have the wrong signs.
Hence the correct relations are (b) and (d).