Answer
$7.45\;\rm kg/m^3$
Work Step by Step
To find the mass density of the gas, we need to find its total mass, and to find its total mass, we need to find how many moles it has.
And to find the number of moles, we need to find the gas absolute pressure. And here we go...
We know that the absolute pressure of the neon gas inside the cylinder is given by
$$P_{}=P_{\rm gauge}+P_a$$
Plugging the known;
$$P=P_{\rm gauge}+P_a=\left(120\times \dfrac{1.013\times 10^5}{14.7}\right)+(1.013\times 10^5)$$
$$P=\bf 9.282\times 10^5\;\rm Pa\tag 1$$
We know, for an ideal gas, that
$$PV=nRT$$
Hence,
$$n=\dfrac{PV}{RT} $$
Recalling that
$n=m/M$ where $m$ is the mass of the gas sample and $M$ is atomic mass of the gas' element.
Hence,
$$\dfrac{m}{M_{\rm Neon}}= \dfrac{PV}{RT}$$
Thus,
$$ m= \dfrac{PVM_{\rm Neon}}{RT}\tag 2$$
We know that the mass density is given by
$$\rho=\dfrac{m}{V}$$
So, we need to divide both sides of (2) bt $V$;
$$ \dfrac{m}{V}= \dfrac{P \color{red}{\bf\not} VM_{\rm Neon}}{ \color{red}{\bf\not} VRT}$$
$$ \rho = \dfrac{P M_{\rm Neon}}{RT} $$
Plugging the known from the given, the periodic table of elements, and (1);
$$\rho =\dfrac{( 9.282\times 10^5) (20.2\times 10^{-3} )}{(8.31)(30+273)}$$
$$\rho =\color{red}{\bf 7.45} \;\rm kg/m^3$$
