Kepler had no idea why, but he saw it in the data: for everything orbiting one central body, $T^2$ divided by $a^3$ is the same number. Compare two orbits and that shared number cancels, leaving $\dfrac{T_1^2}{T_2^2} = \dfrac{a_1^3}{a_2^3}$.

The symbols, and the subscript convention stated once and for all: $T_1$ and $T_2$ are the two periods, $a_1$ and $a_2$ the two orbit sizes — the semi-major axis, which for a circle is just the radius. Subscript 1 is the body you are solving for; subscript 2 is the one you already know. Both bodies must circle the SAME central mass, and because every term is a ratio, the units simply have to match each other — days against days, kilometres against kilometres, no conversion required. That is the bargain: you may answer without ever knowing the planet's mass, or G, or anything else.

Working form: $T_1 = T_2\left(\dfrac{a_1}{a_2}\right)^{3/2}$, and backwards $a_1 = a_2\left(\dfrac{T_1}{T_2}\right)^{2/3}$. Four times wider is eight times slower; nine times wider is twenty-seven times slower. The exponents are the lesson — swap them and the answer is not slightly wrong, it is a different universe.