A two-digit number is such that the product of its digits is 14.

Let the digits at units and tens places be $x$ and $y$, respectively.

$\therefore x y=14$

$\Rightarrow y=\frac{14}{x} \quad \ldots(\mathrm{i})$

According to the question:

$(10 y+x)+45=10 x+y$

$\Rightarrow 9 y-9 x=-45$

$\Rightarrow y-x=-5$                     ........(ii)

From (i) and (ii), we get:

$\frac{14}{x}-x=-5$

$\Rightarrow \frac{14-x^{2}}{x}=-5$

$\Rightarrow 14-x^{2}=-5 x$

$\Rightarrow x^{2}-5 x-14=0$

$\Rightarrow x^{2}-(7-2) x-14=0$

$\Rightarrow x^{2}-7 x+2 x-14=0$

$\Rightarrow x(x-7)+2(x-7)=0$

$\Rightarrow(x-7)(x+2)=0$

$\Rightarrow x-7=0$ or $x+2=0$

$\Rightarrow x=7$ or $x=-2$

$\Rightarrow x=7 \quad(\because$ the digit cannot be negative $)$

Putting $x=7$ in equation (i), we get:

$y=2$

$\therefore$ Required number $=10 \times 2+7=27$