Step 1: Write the moment-transfer relation as two equations.
Using $c_{m,ref} = c_{m,ac} + c_l(h - h_{ac})$ for two of the given data points, with $h = 0.3$:
At $c_l = 0.2$: $-0.02 = c_{m,ac} + 0.2(0.3-h_{ac})$
At $c_l = 0.8$: $0.04 = c_{m,ac} + 0.8(0.3-h_{ac})$
Step 2: Eliminate $c_{m,ac}$ by subtracting the equations.
$0.04-(-0.02) = (0.8-0.2)(0.3-h_{ac})$
$0.06 = 0.6(0.3-h_{ac})$
$0.3-h_{ac} = 0.1$
Step 3: Solve for $h_{ac}$.
$h_{ac} = 0.3-0.1 = 0.2$
Step 4: Check with the other data points.
Substituting back, $c_{m,ac} = -0.02 - 0.2(0.1) = -0.04$. Checking with $c_l=0.4$: $c_{m,ref} = -0.04+0.4(0.1) = 0$, which matches the table, confirming the answer.
Final Answer:
\[ \boxed{h_{ac} = 0.2} \]