Simple and direct answer (tldr)
In $1/\rho$ the number 1 is not an uncertain number, it is an exact number with unlimited significant figures. So the significant figures is the same as for your value of $\rho$, which is 2. So $$1/\rho = 1.4 \times 10^6$$
Problem with significant figures here
Significant figures are not used in actual science. They are like "training wheels" for students, so that they can focus on other things. This example is a good one showing why they are not used.
The statement $\rho = 7.4 \times 10^{-7}$ basically means that $\rho$ is somewhere in the interval $$\rho = [7.35 \times 10^{-7},7.45 \times 10^{-7}]$$ Similarly, the statement $1/\rho = 1.4 \times 10^6$ means that $1/\rho$ is somewhere in the interval $$1/\rho = [1.35 \times 10^6, 1.45 \times 10^6]$$
However, if we take the endpoints of $\rho = [7.35 \times 10^{-7},7.45 \times 10^{-7}]$ and transform them, instead of the range implied from significant figures, we get a more correct range of $$1/\rho = [1.34 \times 10^6,1.36 \times 10^6]$$
So the range implied by significant figures does not well represent our actual uncertainty range. The significant figures vastly overstate our final uncertainty, and there is a considerable shift so that some values that we expect would be outside of the range suggested by the significant figures and other values that are suggested by the significant figures are quite far out of the actual range that we expect.
Correct answer that fixes the problem
So, to represent actual uncertainty in science we use a different approach. Instead of just stating a value to a certain number of significant figures, we explicitly state the value and its standard uncertainty. So if we assume that the range $\rho = [7.35 \times 10^{-7},7.45 \times 10^{-7}]$ represents a 95% confidence that the value is in that range, then that corresponds to a standard uncertainty of $u_\rho = 0.025 \times 10^{-7}$ (because a 95% confidence interval is roughly $\pm$ two standard deviations) so we would write it in any of the following formats $$\rho = 7.400 \times 10^{-7} \pm 0.025 \times 10^{-7}$$$$\rho = (7.400\pm 0.025) \times 10^{-7}$$$$\rho = 7.400(25) \times 10^{-7}$$ I personally prefer the middle format, so I will use that for the remainder of this answer.
To calculate the uncertainty in the inverse we use the propagation of uncertainty formula. Here $f(\rho) = 1/\rho$ and the uncertainty in $f(\rho)$ is given by $$u_{f(\rho)}=\sqrt{\left(\frac{\partial f}{\partial \rho} \right)^2 u_\rho{}^2}=\sqrt{\frac{u_\rho{}^2}{\rho^4}}=0.0046 \times 10^6$$ Traditionally we report two digits of uncertainty. So we would write $$1/\rho = (1.3514 \pm 0.0046) \times 10^6$$ This means that with about 95% confidence we would expect $1/\rho$ to be in the range $1/\rho = [1.3422 \times 10^6,1.3605 \times 10^6]$ which is much more consistent with what we got by transforming the endpoints of the range than the naïve use of significant figures would suggest.
