Answer
$78.5\;\rm dB$
Work Step by Step
First, we need to sketch the problem as seen below.
We need to find $\beta_2$ where $\beta_1=75\;\rm dB$, so
$$\Delta \beta=\beta_2-\beta_1$$
$$\Delta \beta=10\log_{10}\left(\dfrac{I_2}{I_0}\right)-10\log_{10}\left(\dfrac{I_1}{I_0}\right)$$
$$\Delta \beta=10\left[\log_{10}\left(\dfrac{I_2}{I_0}\right)- \log_{10}\left(\dfrac{I_1}{I_0}\right)\right]$$
$$\Delta \beta=10 \log_{10}\left(\dfrac{\dfrac{I_2}{I_0}}{\dfrac{I_1}{I_0}}\right) $$
Thus,
$$\beta_2-\beta_1=10 \log_{10}\left(\dfrac{ I_2}{I_1}\right) $$
$$\beta_2=10 \log_{10}\left(\dfrac{ I_2}{I_1}\right)+\beta_1\tag 1 $$
Now since the two loudspeakers broadcast equally in all directions,
$$I=\dfrac{P}{A}$$
where $P$ is the power emitted by the source, and $A=4\pi R^2$ is the surface area of a sphere.
Hence,
$$I_1=\dfrac{P_{S_1}}{4\pi \left(\frac{L}{2}\right)^2}+\dfrac{P_{S_2}}{4\pi \left(\frac{L}{2}\right)^2}$$
where $P_{S_1}=P_{S_2}=P$;
$$I_1=\dfrac{P }{ \pi L^2}+\dfrac{P}{ \pi L^2}$$
$$I_1= \dfrac{2P}{ \pi L^2}\tag 2$$
And
$$I_2=\dfrac{P_{S_1}}{4\pi \left(\frac{3L}{4}\right)^2}+\dfrac{P_{S_2}}{4\pi \left(\frac{L}{4}\right)^2}$$
$$I_2=\dfrac{4P }{ 9\pi L^2}+\dfrac{4P}{ \pi L^2}$$
$$I_2=\dfrac{40P }{ 9\pi L^2} =\dfrac{20 }{ 9 }\left[\dfrac{2P }{ \pi L^2}\right]$$
Plugging from (2);
$$I_2 =\dfrac{20 }{ 9 } I_1$$
Thus,
$$\dfrac{I_2}{I_1}=\dfrac{20 }{ 9 }$$
Plugging into (1) and plugging the known;
$$\beta_2=10 \log_{10}\left(\dfrac{ 20}{9}\right)+75 $$
$$\beta_2=\color{red}{\bf 78.5}\;\rm dB$$
