the value of delta G for formation of cr203 is -540Kj per mole and that of a Al2O3 is -827 kilo joule per mole is a reduction of Cr2o3 possible with aluminium?

Dear Student 

As we know that the more negative the gibbs free energy more spontaneous will be the reaction proceed. 

The Delta G value for $C{r}_2{O}_3$ = - 540 kJ/ mol
The Delta G value for $A{l}_2{O}_3$ = - 827 kJ/ mol

This shows that the formation of $A{l}_2{O}_3$ is more spontaneous. 
Hence on the reaction of $C{r}_2{O}_3$ with Al it easily reduce $C{r}_2{O}_3$ into Cr and itself get oxidized to form $A{l}_2{O}_3$

It can be understood by different way as the $∆G^{°} = -nF{E}^{°}$.
And the reaction is 

$$\begin{gathered} C{r}_2{O}_3 +\hspace{0.17em}2Al \to 2Cr +\hspace{0.17em}A{l}_2{O}_3 \\ As ∆G^{°} = -nF{E}^{°} \\ {E^{°}}_{(Cr/C{r}_2{O}_3) }= -\frac{-540 \times 1000}{2 \times 96485} =2.798 V \\ {E^{°}}_{(C{r}_2{O}_3/Cr)} = -2.798 V \\ {E^{°}}_{(Al/A{l}_2{O}_3) }= -\frac{-827 \times 1000}{2 \times 96485} =4.285 V \\ {E^{°}}_{(A{l}_2{O}_3/Al)} = -4.285 V \\ {E^{°}}_{cell }= E_{cat}^{red} - E_{anode}^{red} \\ =-2.798 - (-4.285) \\ = 1.48 V \end{gathered}$$
Hence as the Standard cell potential is positive which shows that the cell reaction is feasible. 
Yes the reduction of $C{r}_2{O}_3$ with Al is possible.

Regards