How do you balance the following redox reaction under acidic condition?

Cr2O7 2- + H2SO3 -> Cr3+ + SO4 2-

Urgent! Thanks

Your equation is not complete. You have missed quite a few things. Also, it's better to consider the half equations first. The equation you posted is not a half equation. A half equation contains electrons.

Cr_2 O_7\,^{2-} \,+\,14 H^+ \,+ \,6e^-\,\rightarrow\, 2Cr^{3+} \,+ \,7H_2O

SO_3\,^{2-} \,+ \,H_2O\, - \,2e^- \,\rightarrow \,SO_4\,^{2-} \,+\,2H^+

Multiply everything by 3 in the second equation to balance the electrons in both equations:

3SO_3\,^{2-} \,+ \,3H_2O\, - \,6e^- \,\rightarrow \,3SO_4\,^{2-} \,+\,6H^+

Now add both:

Cr_2 O_7\,^{2-} \,+\,14 H^+ \,+ \,6e^-\, +\,3SO_3\,^{2-} \,+ \,3H_2O\, - \,6e^- \,\rightarrow\, 2Cr^{3+} \,+ \,7H_2O\, +\,SO_4\,^{2-} \,+\,2H^+

Cross out commons:

Cr_2 O_7\,^{2-} \,+\, ^{\small{12}} \cancel{14} H^+ \,+ \cancel{\,6e^-\,} +\,3SO_3\,^{2-} \,+ \,\cancel{3H_2O}\, - \cancel{\,6e^-} \,\rightarrow\, 2Cr^{3+} \,+ \,^{\small4}\cancel{7}H_2O\, +\,SO_4\,^{2-} \,+\,\cancel{2H^+}

Cr_2 O_7\,^{2-} \,+\,12 H^+ \,+ \,3SO_3\,^{2-} \,\rightarrow\, 2Cr^{3+} \,+ \,4H_2O\, +\,SO_4\,^{2-}