It would take $6$ time units.
We know that:
$G_i = A_iB_i,$
$P_i = A_i\oplus B_i$ and
$S_i = P_i\oplus C_i$
$C_1 = G_0 + P_0C_0$
$C_2 = G_1 + P_1G_0 + P_1P_0C_0$
$C_3 = G_2 + P_2G_1 + P_2P_1G_0 + P_2P_1P_0C_0$
$C_4 = G_3 + P_3G_2 + P_3P_2G_1 + P_3P_2P_1G_0 + P_3P_2P_1P_0C_0$
XOR can be implemented in 2 levels; level-1 ANDs and Level-2 OR. Hence it would take 2 time units to calculate $P_i$ and $S_i$
The 4-bit addition will be calculated in 3 stages
1. (2 time units) In 2 time units we can compute $G_i$ and $P_i$ in parallel. 2 time units for $P_i$ since its an XOR operation and 1 time unit for $G_i$ since its an AND operation.
2. (2 time units) Once $G_i$ and $P_i$ are available, we can calculate the caries, $C_i$, in 2 time units.
Level-1 we compute all the conjunctions (AND). Example $P_3G_2, P_3P_2G_1, P_3P_2P_1G_0$ and $P_3P_2P_1P_0C_0$ which are required for $C_4$.
Level-2 we get the carries by computing the disjunction (OR).
3. (2 time units) Finally we compute the Sum in 2 time units, as its an XOR operation.
Hence, the total is 2 + 2 + 2 = 6 time units.