Step 1: Set up the rule.
In a neutral molecule all oxidation numbers add up to zero. We use the known numbers for hydrogen and oxygen to find the one for sulphur.
Step 2: Fix the known values.
In $\text{H}_2\text{SO}_4$ each hydrogen is $+1$ and each oxygen is $-2$. Let the oxidation number of sulphur be $x$.
Step 3: Count each atom.
There are 2 hydrogens, 1 sulphur and 4 oxygens. So the sum is\[ 2(+1) + x + 4(-2) = 0 \]
Step 4: Simplify the numbers.
\[ 2 + x - 8 = 0 \]\[ x - 6 = 0 \]
Step 5: Solve for x.
\[ x = +6 \]So sulphur has an oxidation number of $+6$ here.
Step 6: Final choice.
The oxidation number of sulphur in sulphuric acid is $+6$.\[ \boxed{+6} \]