Answer
$0.49\;\rm mm$
$\rm \approx \;2\;m$
Work Step by Step
First, we need to find this wire's resistance when it dissipates 7.5 W when connected to a 1.5-V battery.
$$P=\dfrac{V^2}{R}$$
and since the wire is connected to the battery, it must have the same voltage.
Hence,
$$R=\dfrac{V_B^2}{P}\tag 1$$
We also know that $R$ is given by
$$R=\dfrac{\rho L}{A}$$
where $A=\pi r^2=\pi (D/2)^2$ where $D$ is the diameter.
$$R =\dfrac{4\rho_{\rm Al}L}{\pi D^2}\tag 2$$
Now we have 2 unknowns ($D$ and $L$) and one equation.
We know that the mass density is given by
$$(\rho_{\rm Al})_{\rm mass}=\dfrac{m}{V}$$
Hence,
$$V=\dfrac{m}{(\rho_{\rm Al})_{\rm mass}}$$
where $V=\pi (D/2)^2L$;
$$\pi D^2 L=\dfrac{4m}{(\rho_{\rm Al})_{\rm mass}}\tag 3$$
Solving (2) for $L$ and plug it into (3),
$$\pi D^2 \dfrac{\pi D^2 R}{4\rho_{\rm Al}}=\dfrac{4m}{(\rho_{\rm Al})_{\rm mass}} $$
Hence,
$$ D^4 =\dfrac{16m\rho_{\rm Al}}{\pi ^2 R(\rho_{\rm Al})_{\rm mass}} $$
Plug $R$ from (1),
$$ D =\sqrt[4]{\dfrac{16m\rho_{\rm Al}P}{\pi ^2 V_B^2(\rho_{\rm Al})_{\rm mass}} }$$
Plug the known;
$$ D =\sqrt[4]{\dfrac{16(1\times 10^{-3})(2.8\times 10^{-8})(7.5)}{\pi ^2 (1.5)^2 (2700)} }$$
$$D=\color{red}{\bf 0.487}\;\rm mm$$
Solving (2) for $D^2$ and plug it into (3),
$$\pi L\dfrac{4\rho_{\rm Al}L}{\pi R}=\dfrac{4m}{(\rho_{\rm Al})_{\rm mass}} $$
$$ L^2 =\dfrac{4m R}{ 4\rho_{\rm Al} (\rho_{\rm Al})_{\rm mass}} $$
Plug $R$ from (1),
$$ L =\sqrt{\dfrac{4m V_B^2}{ 4\rho_{\rm Al} (\rho_{\rm Al})_{\rm mass}P} }$$
Plug the known;
$$ L =\sqrt{\dfrac{4(1\times 10^{-3})(1.5)^2}{ 4(2.8\times 10^{-8})(2700)(7.5)} }$$
$$L=\color{red}{\bf 1.99}\;\rm m$$
