Conversion of molecular structure into Fischer projection for compounds having more than two chiral carbons
To do this exercise, you must need to know how to assign stereo configuration according to Cahn–Ingold–Prelog priority rules, and what Fischer projection represent. First, get correct assignment of three stereo centers in the given molecule. That should be $(1R,2R,3S)$ as depicted in the image:

Now, sketch a Fischer projection with three stereo centers without assigning any groups but number the stereo centers as $1,2,3$. Put the least priory group of stereo center 1 (that would be $\ce{CH3}$ group) on top of the vertical line. By definition, top and bottom vertical lines are wedges-down (dashes) while all horizontal lines are wedges-up (solid wedges). If you put highest priority group ($\ce{Cl}$ group) on left horizontal line and second priority group ($\ce{OH}$ group) on right horizontal line, you get the correct clock-wise rotation for $(1R)$ configuration (See the image). You can follow same procedure or the one you are familiar to assign other stereo centers correctly. For example, for stereo center 3, the least priority group is $\ce{H}$ while the highest one is $\ce{F}$ group. The second priority group is substituted carbon chain, which is already in the projection (the last priority to consider is $\ce{CH3}$ group). Thus, I put $\ce{H}$ group on bottom vertical line (should be a dash) as did for stereo center 1. Then, to get correct counter-clock-wise rotation to give $(3S)$ configuration, you must put $\ce{F}$ group on right horizontal line so last priory group $\ce{CH3}$ would occupy the remaining left horizontal point.
Note: You can exchange any group at a stereo center with another group at same center to reverse the configuration, e.g., $(R) \rightarrow (S)$. If you repeat same with two different group, you get back the original configuration. Once I was done with assigning the groups, I have performed these operation twice to change the positions of three groups at stereo center 3 to my likeness. Still, the configuration remains same.
Finally, you may able to assign remaining two groups ($\ce{Br}$ and $\ce{F}$ groups) for stereo center 2 to get $(2R)$ configuration.
To get the enantiomer, you can sketch the mirror image of the compound, which is $(1S,2S,3R)$ as depicted in the image and also tally with the given answer for the corresponding enanthiomer.