 Solution

For balanced global error $\epsilon$, a method second order in both variables can take $h\sim {\epsilon }^{1/2}$ and $k\sim {\epsilon }^{1/2}$. With $O(1/h)$ work per time step in one spatial dimension, its cost scales as ${\epsilon }^{-1}$. A method second order in space but first order in time instead needs $k\sim \epsilon$, so its cost is $\boxed{O({\epsilon }^{-3/2})}$ under the same per-step cost assumption. Equivalently at fixed $h$, using $k\sim h^2$ rather than $k\sim h$ needs about $1/h$ more steps. A restrictive stability condition or different linear-solver cost can dominate these accuracy-based estimates.